If you expand this on both sides of the equation you will find that you are left with a quadratic (there are a number of nice cancellations) that you can solve for x. Then put this value of x into the y equation to find a value of y. (Make sure that the final value you claim is the answer is, in fact, a minimum not a maximum.)

A quicker way to do this is logical. You are adding two positive terms to find a minimum, so find the minimum values of and . Then add these to get the minimum value for y.

may i know what theorem or statment should be used to explain that the stationary is minimum point instead of maximum or inflection point. thank

In general there isn't one. Personally I would use the characteristics of the graph of the function to decide if we have a minimum point there, but in general you need to use the second derivative test:

If then the stationary point is a relative maximum.

If then the stationary point is a relative minimum.

If then the stationary point is a point of inflection.

I mentioned that we need to check this near the bottom of my first post.

Let me add something.--->Inflection point.
But you know nothing about whether it is maximum of minimum! You need to return back to the first derivative test. A common error is that people think whenever the second derivative is zero it means it is neither a minimum nor a maximum (I think the name of the term is 'saddle point')

your explanation of the situation and especially your sketch tickled my memory: This problem was solved by Archimedes already - if I remember correctly. (Reflect R at the x-Axis. The straight line between Q and R' isthe shortest distance between Q and R. P is the zero of this line)